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Homework Help: Multiplying a fraction

  1. Dec 24, 2015 #1
    1. The problem statement, all variables and given/known data
    I would just like to know if I got this right. ((x2 + 2x + 1)/(18cw3) times ((12c3w)/(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3)(x2 - 18cw3

    2. Relevant equations
    ((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))

    3. The attempt at a solution
    (12c3wx2 + 2x12c3w + 12c3w)/((18cw3)(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3x2 - 18cw3
     
  2. jcsd
  3. Dec 24, 2015 #2

    Samy_A

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    Removed, as apparently there is an error in the formulation of the question.
     
  4. Dec 24, 2015 #3
    I accidentally used a subscript all the way through instead of a power, argh!!
     
  5. Dec 24, 2015 #4

    Mark44

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    On the right side one of the terms is ##2x12c_3w##. Should that have been ##2x^{12}c^3##? Also, you're missing a right parenthesis.
     
  6. Dec 25, 2015 #5

    haruspex

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    Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?
    If so, there is a lot of cancellation available. Factorise the two expressions involving x.
     
  7. Dec 25, 2015 #6
    It should have been as haruspex said: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1)
     
  8. Dec 25, 2015 #7
    Yes, that is what I meant.
     
  9. Dec 25, 2015 #8
    (x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)
    The x+1 cancels, leaving: (x+1)/(18cw3) times (12c3w)/(x-1)
     
  10. Dec 25, 2015 #9

    haruspex

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    More cancellation to go. Look at the constants, look at the powers of c and w.
     
  11. Dec 25, 2015 #10
    (x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)
     
  12. Dec 25, 2015 #11
    The 3w is canceled
     
  13. Dec 25, 2015 #12
    I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)
     
  14. Dec 25, 2015 #13

    haruspex

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    You made a mistake with the constants.
     
  15. Dec 25, 2015 #14
    Should it be: 3cw(w2) and 3cw(4c2)??
     
  16. Dec 25, 2015 #15
    Oops I meant: 3cw(6w2) and 3cw(4c2)
     
  17. Dec 25, 2015 #16

    haruspex

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    Yes.
     
  18. Dec 25, 2015 #17
    Thank you for your help
     
  19. Dec 25, 2015 #18
    So the answer is: (x+1)/(6w2) times (4c2)/(x-1)
     
  20. Dec 26, 2015 #19

    Mark44

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    Instead of adding new posts each time you discover something you should have said, you can edit the post you want to change.
     
  21. Dec 26, 2015 #20
    The constants can be further simplified, and then you should bring it together so you have one fraction.
     
  22. Dec 26, 2015 #21
    Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?
     
  23. Dec 27, 2015 #22

    Samy_A

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    1)
    2) I think you can leave the nominator as ##2c²(x+1)##, and similarly for the denominator. Certainly would give a better readable result.
     
    Last edited: Dec 27, 2015
  24. Dec 27, 2015 #23
    How did you get 2c2(x+1) when it was 4c2/(x+1)?? Isn't the numerator supposed to be 2c2(2x+2)??
    And the denominator is supposed to be: 2w2(3x-3)??
     
  25. Dec 27, 2015 #24

    haruspex

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    You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.
     
  26. Dec 27, 2015 #25
    Okay, I think I have it now: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1) = (x+1)2(12c3w)/(18cw2)(x+1)(x-1) Cancel the (x+1) = (12c3w)(x+1)/(18cw2)(x-1) Cancel the c and w = (12c2)(x+1)/(18w)(x-1) divided by 3 (numerator and denominator) = (4c2)(x+1)/(6w)(x-1) divided by 2 (numerator and denominator) = (2c2)(x+1)/(3w)(x-1) Answer
     
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